The proof repeats almost literally the demonstration of the well-known facts [1]} about the ordinary pointwise convergence. Namely, let \((x_n^*) \subset Y^*\)  be a bounded sequence. Recall, that \(\sigma (Y^*, Y)\) -boundedness is equivalent to the boundedness in norm (Banach-Steinhaus), so \(\sup _n\Vert x_n*\Vert  = C < \infty \) . Assume that for some filter \(\mathfrak {F}\)  on \({\mathbb {N}}\)  the sequence \((x_n^*)\)  is \(\mathfrak {F}\) -Cauchy in topology \(\sigma (Y^*, Y)\) . This means that for every \(x \in Y\)  the sequence \((x_n^*(x)) \subset {\mathbb {K}}\)  is \(\mathfrak {F}\) -Cauchy. By completeness of \({\mathbb {K}}\)  (which is either \({\mathbb {R}}\)  or \(), \) F xn*(x)\( exists for all \) xY\(. Consider the map \) f Y K\( given by the recipe \) f(x) = F xn*(x)\(. At first, it is a linear functional. Indeed,\)  f(ax1 + bx2) = F xn* (ax1 + bx2) = a F xn* (x1) + b F xn* (x2) = af(x1) + f(x2)\(.At second, the estimate \) |f(x)| = F xn*(x) Cx\(, which holds for all \) xY\(, demonstrates that \) f\( is continuous, so \) f Y*\(, and \) f = F xn*\( in \) (Y*, Y)\(.\)
Theorem 4.9 
Let \(Y\)  be a Banach space, \(\mathfrak {F}\)  be a stationary B-UBP-filter on \({\mathbb {N}}\) , then \((Y^*, \sigma (Y^*, Y) )\)  is \(\mathfrak {F}\) -complete.

Let \(x^* = (x_n^*) \subset Y^*\)  be \(\mathfrak {F}\) -Cauchy in the topology \(\sigma (Y^*, Y)\) . Then for every \(x \in Y\)  the sequence \((x_n^*(x)) \subset {\mathbb {K}}\)  is \(\mathfrak {F}\) -Cauchy, which implies that \(x^*\)  is pointwise \(\mathfrak {F}\) -bounded. From the definition of stationary B-UBP-filter we deduce the existence of an \(\mathfrak {F}\) -stationary set \(A \subset {\mathbb {N}}\)  such that \(\sup _{n \in A} \Vert x_n^*\Vert  < \infty \) . Consider the collection \(G\)  all sets of the form \(A\cap B\) , \(B \in \mathfrak {F}\) . \(G\)  is a filter base. Denote \(\mathfrak {F}_A\)  the filter on \({\mathbb {N}}\)  generated by this particular base \(G\) . Let \(g : {\mathbb {N}}\rightarrow A\)  be a bijection. Denote \(\mathfrak {F}_g\)  the filter of all those \(B \subset {\mathbb {N}}\)  for which \(g(B) \in G\) . Finally, consider \(y^* = x^* \circ g\) . Then \(y^* \)  is pointwise bounded and is \(\mathfrak {F}_g\) -Cauchy. By Theorem REF  the sequence \(y^*\)  is pointwise \(\mathfrak {F}_g\) -convergent to some \(f \in Y^*\) . This means that the sequence \(x^*\)  converges to \(f\)  with respect to the filter \(g[\mathfrak {F}_g] = F_A\) . By the construction, \(\mathfrak {F}_A \supset \mathfrak {F}\) , consequently, \(f\)  is an \(\mathfrak {F}\) -cluster point for \(x^* = (x_n^*)\)  in \(\sigma (Y^*, Y)\) . But \(x^*\)  is \(\mathfrak {F}\) -Cauchy, so its cluster point \(f\)  is its limit (Proposition REF ).
We don't know whether for general TVS the sequential completeness implies \(f\) -statistical completeness for every unbounded modulus \(f\) , but we have an analogous result for bounded completeness in locally convex spaces.
Recall that in the particular case of the modulus function \(f(t) = t\) , \(f\) -statistical convergence reduces to the well-known statistical convergence, which is generated by the filter \(\mathfrak {F}_{st}\) , whose elements are those \(A \subset {\mathbb {N}}\) , for which
\( \lim _{n \rightarrow \infty } \frac{ |A(n)|}{n} = 1,\) 
where \(A(n) = A \cap \overline{1, n}\) . First, we need a generalization of the following fact that was remarked already in [2]}: if a bounded numerical sequence \((x_n)\)  converges statistically to a number \(a\) , then it is Cesaro convergent to \(a\) , i.e.
\(\lim _{n \rightarrow \infty } \frac{1}{n}\sum _{k=1}^n x_k = a.\) 
Lemma 4.10 
Let \(X\)  be a locally convex TVS and \(x = (x_n) \in 2^X\)  be a bounded \(\mathfrak {F}_{st}\) -Cauchy sequence in \(X\) , then the sequence \(y = (y_n) \in 2^X\) , where \(y_n = \frac{1}{n}\sum _{k=1}^n x_k\) , is a bounded Cauchy sequence.

Let \(U \in \mathfrak {N}_0\)  be an open balanced convex neighborhood of zero and \(p\)  be the seminorm whose open unit ball is equal to \(U\) . Denote \(C = \sup _n p(x_n)\) . Then all \(x_k \in C U\)  and, by convexity, all \(y_n \in C U\) , which proves the boundness of \(y\) . It remains to show that \(y\)  is a Cauchy sequence.
According to our assumption, there is a set \(A \subset {\mathbb {N}}\)  that satisfies (REF ) and such that \(p(x_n - x_m) < \frac{1}{2}\)  for all \(m, n \in A\) 
Select an \(N \in {\mathbb {N}}\)  in such a way that for all \(n \geqslant N\)
\(\frac{ |({\mathbb {N}}\setminus A)(n)|}{n} < \frac{1}{8C}.\) 
Then, for \(n, m \geqslant N\)  we have
\(&p\left( y_n - y_m \right) = p\left( \frac{1}{n}\sum _{k=1}^n x_k - \frac{1}{m}\sum _{j=1}^m x_j \right) = p\Bigl ( \frac{1}{n}\sum _{k \in ({\mathbb {N}}\setminus A) (n)} x_k \\&- \frac{1}{m}\sum _{j \in ({\mathbb {N}}\setminus A) (m)} x_j + \frac{1}{|A(n)| |A(m)|}\sum _{k \in A(n), \ j \in A (m)} (x_k - x_j) \\&- \left(\frac{1}{|A(n)| } - \frac{1}{n} \right)\sum _{k \in A(n)}x_k - \left(\frac{1}{|A(m)| } - \frac{1}{m} \right)\sum _{j \in A(m)}x_j)\Bigr )\) 
\(&\leqslant \frac{1}{n}\sum _{k \in ({\mathbb {N}}\setminus A) (n)} p(x_k)+ \frac{1}{m}\sum _{j \in ({\mathbb {N}}\setminus A) (m)} p(x_j) \\&+ \frac{1}{|A(n)| |A(m)|}\sum _{k \in A(n), \ j \in A (m)} p(x_k - x_j) \\&+ \left(\frac{1}{|A(n)| } - \frac{1}{n} \right)\sum _{k \in A(n)}p(x_k) + \left(\frac{1}{|A(m)| } - \frac{1}{m} \right)\sum _{j \in A(m)}p(x_j) \\& < \frac{C}{n} \frac{n}{8C} + \frac{C}{m} \frac{m}{8C} + \frac{1}{2} + \frac{C}{n} \frac{n}{8C} + \frac{C}{m} \frac{m}{8C} = 1,\) 

that is \(y_n - y_m \in U\) .
Theorem 4.11 
Let \(X\)  be a boundedly sequentially complete locally convex TVS, then \(X \in {\text{Compl}}_b(\mathfrak {F}_{st})\) .

Let \(x = (x_n) \in 2^X\)  be a bounded \(\mathfrak {F}_{st}\) -Cauchy sequence in \(X\) . According to the previous lemma, the sequence \(y = (y_n) \in 2^X\) , where \(y_n = \frac{1}{n}\sum _{k=1}^n x_k\) , is a bounded Cauchy sequence, so it has a limit in \(X\) .
Denote \(a = \lim _{n \rightarrow \infty } y_n \in X\) . Let us demonstrate that \(a = \lim _{\mathfrak {F}_st} y\) . To do this, consider \(U\) , \(p\) , \(C\) , \(A\)  and \(N\)  from the proof of Lemma REF . Also, fix such an \(M > N\)  that \(p(y_n - a) < \frac{1}{4}\)  for all \(n > M\) . Then, for every \(n \in A \setminus \overline{1, M}\)  we have
\(p(x_n - a) \leqslant \frac{1}{4} + p(x_n - y_n) \leqslant \frac{1}{4} + p\left(\frac{1}{n}\sum _{k \in ({\mathbb {N}}\setminus A) (n)} x_k \right) + p\left( x_n - \frac{1}{n}\sum _{k \in A(n)} x_k \right)\) 
\(&\leqslant \frac{1}{4} + \frac{C}{n} \frac{n}{8C} + p\left( \frac{1}{|A(n)|}\sum _{k \in A(n)} (x_n - x_k ) \right)+ \left(\frac{1}{|A(n)|}- \frac{1 }{n} \right) p\left(\sum _{k \in A(n)} x_k \right) \\&\leqslant \frac{1}{4} + \frac{C}{n} \frac{n}{8C} + \frac{1}{2} + \frac{C}{n} \frac{n}{8C} = 1,\) 

that is \(x_n - a \in U\) .
Corollary 4.12 
Let \(X\)  be a boundedly sequentially complete locally convex TVS, then \(X \in {\text{Compl}}_b(\mathfrak {F}_{f-st})\)  for every unbounded modulus function \(f\) .

Since \(\mathfrak {F}_{f-st} \subset \mathfrak {F}_{st}\)  (see the reasoning just before [3]}), it remains to apply the statement (4) of theorem REF  in its version for bounded sequences, which works the same way as the original one.

Completeness and ultrafilters
Let us start with an easy observation.
Remark 4.13 
If a TVS \(X\)  is complete with respect to all ultrafilters on \({\mathbb {N}}\) , then \(X\)  is countably complete. Select an ultrafilter \(\mathfrak {U}\supset \mathfrak {F}\) . By our assumption, \(X \in {\text{Compl}}(\mathfrak {U})\) , and it remains to apply (4) of Theorem REF  in order to show that \(X \in {\text{Compl}}(\mathfrak {F})\) .

The above remark motivates some natural questions. At first, is it true that the completeness with respect to one ultrafilter implies the completeness with respect to all other ultrafilters (and hence implies the countable completeness)? If the answer is negative, then the second question arises: does the sequential completeness imply completeness with respect to some ultrafilter? The negative answers to both questions are given below (for the first one the answer is given under an additional set-theoretic assumption).
Theorem 4.14 
Under the Martin's axiom there are free ultrafilters \(\mathfrak {U}_1, \mathfrak {U}_2\)  on \({\mathbb {N}}\)  and a TVS \(X\)  such that \(X \in {\text{Compl}}(\mathfrak {U}_1)\) , but \(X \notin {\text{Compl}}(\mathfrak {U}_2)\) .

Let \(X = (Y^*, \sigma (Y^*, Y) )\) , where \(Y\)  is a separable infinite-dimensional Banach space. According to [4]}, the Martin's axiom guaranties the existence of
\(2^{2^{\aleph _0}}\) 
-many B-UBP-ultrafilters. Let \(\mathfrak {U}_1\)  be a B-UBP-ultrafilter on \({\mathbb {N}}\) . Due to Theorem REF , \(X \in {\text{Compl}}(\mathfrak {U}_1)\) . On the other hand, \(X\)  is separable (the dual to a separable Banach space contains a countable total system [1]} and, consequently, is w\(^*\) -separable) and incomplete, so by Theorem REF  \(X\)  is not countably complete, which implies (Remark REF ) that \(X \notin {\text{Compl}}(\mathfrak {U}_2)\)  for some free ultrafilter \(\mathfrak {U}_2\)  on \({\mathbb {N}}\) .
Recall, that the dual to \(\ell _1\)  is \(\ell _\infty \) , and for every \(x = (x_1, x_2, \ldots ) \in \ell _\infty \)  and \(y = (y_1, y_2, \ldots ) \in \ell _1\)  the action of \(x\)  on \(y\)  is \(x(y) = \sum _{n \in {\mathbb {N}}}x_n y_n\) .
Theorem 4.15 
The space \(\ell _1\)  in the weak topology \(\sigma (\ell _1, \ell _\infty )\)  is sequentially complete, but is not boundedly complete over any free ultrafilter \(\mathfrak {U}\)  on \({\mathbb {N}}\) .

Weak sequential completeness of \(\ell _1\)  (as well as of all spaces \(L_1(\mu )\) ) is a classical Banach space theory result, see [6]}. Now, let us fix an arbitrary free ultrafilter \(\mathfrak {U}\)  on \({\mathbb {N}}\)  and demonstrate that \((\ell _1, \sigma (\ell _1, \ell _\infty )) \notin {\text{Compl}}_b(\mathfrak {U})\) . Denote \(\left\lbrace {e_n} \right\rbrace _1^\infty \)  the canonical basis of \(\ell _1\) , that is \(e_1 = (1,0,0,\ldots )\) , \(e_2 = (0,1,0,\ldots )\) ,.... For every \(x = (x_1, x_2, \ldots ) \in \ell _\infty \)  the values \(x(e_n) = x_n\)  form a bounded sequence of scalars, hence there is the limit on \(x(e_n)\)  over \(\mathfrak {U}\) . Consequently, \((x(e_n))\)  is \(\mathfrak {U}\) -Cauchy in the topology \(\sigma (\ell _1, \ell _\infty )\) .
Now we show that the sequence \((e_k)\)  does not have a weak limit over \(\mathfrak {U}\) . Assume that there exists \(z =(z_1, z_2, \ldots ) \in \ell _1\)  such that \(x(z) = \lim _\mathfrak {U}(x(e_n)) = \lim _\mathfrak {U}(x_n)\)  for all \(x = (x_1, x_2, \ldots ) \in \ell _{\infty }\) . Then, on the one hand, considering \(e_k\)  as elements of \(\ell _\infty \)  we get that for every \(k \in {\mathbb {N}}\)
\(z_k = e_k(z) = \lim _{\mathfrak {U}, n} (e_k(e_n)) = 0,\) 
but on the other hand, taking \(x = (1, 1, 1, \ldots )\)  we get that
\(\sum _{n \in {\mathbb {N}}} z_n = \lim _\mathfrak {U}(x_n) = 1\) .
We came to a contradiction.
The above result can be viewed in a bit different way. Consider \(\ell _1\)  as a subspace of \(\ell _\infty ^*\) . Then in \(\sigma (\ell _\infty ^*, \ell _\infty )\)  we have that \((e_k)\)  is \(\mathfrak {U}\) -convergent to the functional \(x \mapsto \lim _\mathfrak {U}x\) , and this functional belongs to \(\ell _\infty ^* \setminus \ell _1\) .

Concluding remarks and open questions
The following challenging problem remains open.
Problem 5.1 
Is there a combinatorial description of those filter \(\mathfrak {F}\)  on \({\mathbb {N}}\)  for which the sequential completeness of a TVS implies its \(\mathfrak {F}\) -completeness?

Problem 5.2 
Which of concrete filters \(\mathfrak {F}\) , widely mentioned in literature (like Erdös-Ulam filters, summable filters, \(f\) -statistical filters, filters generated by summability matrices, etc.) enjoy the property that \(\mathfrak {F}\) -sequential completeness of a TVS implies its \(\mathfrak {F}\) -completeness?

Let us consider the following construction. For a free filter \(\mathfrak {F}\)  on \({\mathbb {N}}\)  denote \(c(\mathfrak {F})\)  the set of all bounded \(\mathfrak {F}\) -convergent numerical sequences. Evidently \(c_0 \subset c(\mathfrak {Fr}) \subset c(\mathfrak {F}) \subset \ell _\infty \) . For an ultrafilter \(\mathfrak {U}\)  we have \(c(\mathfrak {F}) = \ell _\infty \) . Following the argument from Theorem REF  one can easily see that the space \((\ell _1, \sigma (\ell _1, c(\mathfrak {F})))\)  is not \(\mathfrak {F}\) -complete. So, every time when \((\ell _1, \sigma (\ell _1, c(\mathfrak {F})))\)  is sequentially complete, we obtain an example of a sequentially complete space which is not \(\mathfrak {F}\) -complete. This relates Problem REF  the following one.
Problem 5.3  \( \ \)

Describe those linear subspaces \(E\) , \(c_0 \subset E \subset \ell _\infty \) , for which the corresponding space \((\ell _1, \sigma (E)\)  is sequentially complete.

Describe those free filters \(\mathfrak {F}\)  on \({\mathbb {N}}\) , for which the corresponding space \((\ell _1, \sigma (\ell _1, c(\mathfrak {F})))\)  is sequentially complete.


Remark, that for some filters \(\mathfrak {F}\)  the corresponding space \((\ell _1, \sigma (\ell _1, c(\mathfrak {F})))\)  is not sequentially complete. This evidently happens for the the Fréchet filter and for those filters \(\mathfrak {F}\)  for which the implication (\((X \in {\text{Compl}}(\mathfrak {Fr})) \Rightarrow (X \in {\text{Compl}}(\mathfrak {F}))\) ) from Problem REF  holds true. Let us give a more advanced example.
Theorem 5.4 
For every unbounded modulus function \(f\)  the space
\((\ell _1, \sigma (\ell _1, c(\mathfrak {F}_{f-st})))\)  is not sequentially complete.

Denote \(x_n = \frac{1}{n}\sum _{k=1}^n e_k\) , where \(e_k \in \ell _1\)  are the elements of the canonical basis. For every \(y = (y_1, y_2, \ldots ) \in c(\mathfrak {F}_{f-st})\)  we know [3]} that it is statistically convergent to its \(f\) -statistical limit, so it is Cesaro convergent. Consequently
\(y(x_n) = \frac{1}{n}\sum _{k=1}^n y_k \xrightarrow[n \rightarrow \infty ]{} \lim _{\mathfrak {F}_{f-st}} (y_n).\) 
This means that \((x_n)\)  is a Cauchy sequence in \((\ell _1, \sigma (\ell _1, c(\mathfrak {F}_{f-st})))\) . On the other hand, \((x_n)\)  is not \(\sigma (\ell _1, c(\mathfrak {F}_{f-st}))\) -convergent to any element of \(z \in \ell _1\)  because the mapping \(y \mapsto \lim _{\mathfrak {F}_{f-st}} (y_n)\)  cannot be represented in the form \(y \mapsto \sum _{n \in {\mathbb {N}}}z_n y_n\) .
Problem 5.5 
Does there exist a “universal” ultrafilter \(\mathfrak {U}\)  on \({\mathbb {N}}\) , such that completeness with respect to \(\mathfrak {U}\)  implies the countable completeness? Is the existence of such \(\mathfrak {U}\)  consistent with ZFC axioms?

Problem 5.6 
For a given TVS \(X\)  denote \({\text{Compl}}(X)\)  the set of those filters \(\mathfrak {F}\)  for which \(X \in {\text{Compl}}(\mathfrak {F})\) . Items (4–6) of Theorem REF  give some restrictions on the structure of \({\text{Compl}}(X)\) . What else can be said about this set? For example, are there any topological restrictions on the intersection of \({\text{Compl}}(X)\)  with the space \(\beta {{\mathbb {N}}}\)  of all ultrafilters?

Remark that the questions formulated in Problems REF , REF , REF , and REF  can be asked for smaller classes of spaces, for example for locally convex spaces.
